Question: Medians $\overline{AD}$ and $\overline{BE}$ of $\triangle ABC$ are perpendicular.  If $AD= 15$ and $BE = 20$, then what is the area of $\triangle ABC$?
Solution: Let the medians intersect at point $G$ as shown below.  We include the third median of the triangle in red; it passes through the intersection of the other two medians.

[asy]
pair D,EE,F,P,Q,G;

G = (0,0);
D = (-1,0);
P= (0.5,0);
EE = (0,4/3);
Q = (0,-2/3);
F = 2*Q - D;
draw(P--D--EE--F--D);
draw(EE--Q);
label("$A$",D,W);
label("$D$",P,NE);
label("$E$",Q,SW);
label("$B$",EE,N);
label("$C$",F,SE);
draw(rightanglemark(P,G,EE,3.5));
label("$G$",G,SW);
draw(F--(D+EE)/2,red);
[/asy]

Point $G$ is the centroid of $\triangle ABC$, so $AG:GD = BG:GE = 2:1$.  Therefore, $AG = \frac23(AD) = 10$ and $BG = \frac23(BE) = \frac{40}{3}$.

Drawing all three medians of a triangle divides the triangle into six triangles with equal area.  In $\triangle ABC$ above, $\triangle ABG$ consists of two of these six triangles, so the area of $\triangle ABC$ is 3 times the area of $\triangle ABG$: \[ [ABC] = 3[ABG] = 3\cdot \frac12 \cdot AG \cdot BG = \frac32\cdot 10 \cdot \frac{40}{3} = \boxed{200}.\]